Numbers with multiple runs of ones

Question

Task

Find the set of numbers such that the binary representation contains two or more runs of 1 separated by at least one 0.

For example, the for the numbers that are 4 bits long:

 0 0000        (no ones)
 1 0001        (only one run)
 2 0010        (only one run)
 3 0011        (only one run)
 4 0100        (only one run)
 5 0101 Valid
 6 0110        (only one run)
 7 0111        (only one run)
 8 1000        (only one run)
 9 1001 Valid
10 1010 Valid
11 1011 Valid
12 1100        (only one run)
13 1101 Valid
14 1110        (only one run)
15 1111        (only one run)

Input

An integer provided to the application via some input in the range 3 .. 32. This represents the maximum number of bits to count up to.

The input of n indicates that the numbers 0 .. 2n-1 need to be examined.

Output

A delimited (your choice) list of all numbers meeting the criteria. The numbers are to be presented in numeric order. An extra trailing delimiter is acceptable. Data structure enclosures (e.g. [] and similar) are also acceptable.

Example

Input: 3
Output: 5

Input: 4
Output: 5, 9, 10, 11, 13

Input: 5
Output: 5, 9, 10, 11, 13, 17, 18, 19, 20, 21, 22, 23, 25, 26, 27, 29

This is code-golf - the answer with the least amount of bytes wins.

Answer 1

Python, 48

lambda n:[i for i in range(2**n)if'01'in bin(i)]

I had been vastly overthinking this. We just need to check if the binary expansion contains '01'.

For there to be two runs of ones, the one on the right must be preceded by a 0. If there's only one run, there won't be any leading 0's, so that won't happen.

---

Old answer:

lambda n:[i for i in range(2**n)if len(set(bin(i).split('0')))>2]

The Python binary representation works very nicely here. A binary number is written like bin(9)=='0b10110'. Splitting at '0' results in a list of

• Empty strings to the left of the initial 0, between any two consecutive 0's, and to the right of any final 0
• The letter b followed by one or more leading ones
• Runs of 1's that are not leading

The first two categories always exist, but the last one only exists if there is a run one 1's that doesn't contain the leading '1', and so only if there's more than one run of 1's. So, it suffices to check if the list contains more than 2 distinct elements.

Python 3.5 saves 2 chars by unpacking {*_} in place of set(_).

Answer 2

Ruby, 44 40 38 chars

^{crossed out 44 is still regular 44 ;(}

->n{(0..2**n).select{|x|/01/=~'%b'%x}}

An anonymous function (proc, actually) that takes an integer and returns an array.

Uses the regex /10+1/: a 1, at least one 0, and then another 1. @histocrat points out that if 01 is anywhere in the string, there must be a 1 somewhere before it.

Answer 3

Pyth, 12 bytes

f<2r.BT8U^2Q

Try it online.

Idea

The binary representation of any positive number always begins with a run of 1s, possibly followed by other, alternating runs of 0s and 1s. If there are at least three separate runs, two of them are guaranteed to be runs of 1s.

Code

              (implicit) Store the evaluated input in Q.
         ^2Q  Calculate 2**Q.
f       U     Filter; for each T in [0, ..., 2**Q-1]:
    .BT         Compute T's binary representation.
   r   8        Perform run-length encoding.
                This returns a list of character/run-length pairs.
 <2             Discard the trailing two pairs.
                This returns a non-empty array if there are more than 2 runs.
              Keep T if the array was truthy (non-empty).

Answer 4

Julia, 43 41 bytes

n->filter(i->ismatch(r"01",bin(i)),1:2^n)

This creates an unnamed function that accepts an integer and returns an array. It uses histocrats's regex trick (used in Doorknob's answer), where 01 will only match if there's a preceding 1.

Ungolfed:

function f(n::Int)
    # Take the integers from 1 to 2^n and filter them down to
    # only those such that the binary representation of the integer
    # matches the regex /01/.
    filter(i -> ismatch(r"01", bin(i)), 1:2^n)
end

Answer 5

Matlab, 79 68 64 59

The idea is interpreting the binary number as array of zeros and ones, and then calculating the absolute difference between each pair of neighbours. If we have two or more times a difference of 1, then we obviously have a run of two or more ones. Note that this only works if we represent the binary number without leading zeros.

@(n)find(arrayfun(@(k)sum(~~diff(dec2bin(k)+0))>1,1:2^n-1))

Old versions:

k=1:2^input('')-1;k(arrayfun(@(k)sum(~~diff(dec2bin(k)+0))>1,k))

for k=1:2^input('')-1;if sum(~~diff(dec2bin(k)+0))>1;disp(k);end;end

for k=1:2^input('')-1;if sum(~~conv(dec2bin(k)+0,[-1,1],'v'))>1;disp(k);end;end

Answer 6

JavaScript (ES7), 89 85 72 69 62 bytes

Holy cow, creating ranges in JS is not easy. Perhaps it would be shorter with an actual for loop. Nope, I lied; it's actually a bit longer. Oh well. I guess I'll just have to settle for 27 bytes saved. (7 thanks to Mwr247!)

x=>[for(a of Array(1<<x).keys())if(/01/.test(a.toString(2)))a]

Works properly in the latest versions of Firefox, but probably not in any other browser. Try it out:

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Using no input."});addOutput(newoutput,{type:"err",content:"Input is malformed. Using no input."});input=[]}doInterpret(getNewCode(),input,function(data){addOutput(newoutput,data)});doInterpret(getOriginalCode(),input,function(data){addOutput(origoutput,data)});evt.stopPropagation();return false}function onResetBtn(evt){setTextbox(getOriginalCode());origheader.style.display="none";newheader.style.display="none";setOutput(origoutput,"");setOutput(newoutput,"")}function escape(s){return s.toString().replace(/&/g,"&amp;").replace(/</g,"&lt;").replace(/>/g,"&gt;")}window.alert=function(){};window.prompt=function(){};function doInterpret(code,input,cb){var workerCode=interpret.toString()+";function stdout(s){ self.postMessage( {'type': 'out', 'content': s} ); }"+" function stderr(s){ self.postMessage( {'type': 'err', 'content': s} ); }"+" function kill(){ self.close(); }"+" self.addEventListener('message', function(msg){ interpret(msg.data.code, msg.data.input); });";var interpreter=new Worker(URL.createObjectURL(new Blob([workerCode])));interpreter.addEventListener("message",function(msg){cb(msg.data)});interpreter.postMessage({"code":code,"input":input});setTimeout(function(){interpreter.terminate()},1E4)}setTimeout(function(){getOriginalCode();textbox.addEventListener("input",onTextboxChange);testbtn.addEventListener("click",onTestBtn);resetbtn.addEventListener("click",onResetBtn);setTextbox(getOriginalCode())},100);function interpret(code,input){window={};alert=function(s){stdout(s)};window.alert=alert;console.log=alert;prompt=function(s){if(input.length<1)stderr("not enough input");else{var nextInput=input[0];input=input.slice(1);return nextInput.toString()}};window.prompt=prompt;(function(){try{var evalResult=eval(code);if(typeof evalResult=="function"){var callResult=evalResult.apply(this,input);if(typeof callResult!="undefined")stdout(callResult)}}catch(e){stderr(e.message)}})()};</script>

^{(Snippet taken from this page)}

Suggestions welcome!

Answer 7

Haskell, 68 61 53 bytes

Improvement from Damien

g x|x`mod`4==1=x>4|2>1=g$x`div`2
a x=filter g[1..2^x]

History:

This fixes the bug(Switched == and =, and square instead of power of two). And replace true with 2>1 and false with 1>2. Also thanks to point out that 2^x is always fail. Thanks to Thomas Kwa and nimi

g x|x<5=1>2|x`mod`4==1=2>1|2>1=g$x`div`2
a x=filter g[1..2^x]

Originally

g x|x<5=False|x`mod`4=1==True|2>1=g$x`div`2
a x=filter g[1..(x^2-1)]

If it have to be full program,

g x|x<5=False|x`mod`4==1=True|2>1=g$x`div`2
main=interact$show.a
a x=filter g[1..2^(read x)]

Answer 8

APL, 34 27 bytes

{0~⍨{⍵×2<+/2≢/⍵⊤⍨⍵/2}¨⍳2*⍵}

This creates an unnamed monadic function that accepts an integer on the right and returns an array.

Explanation:

                     }¨⍳2*⍵}  ⍝ For each integer from 1 to 2^input...
              ⍵⊤⍨⍵/2         ⍝ Get the binary representation as a vector
           2≢/                ⍝ Pairwise non-match, yielding a boolean vector
       2<+/                   ⍝ Check whether the number of trues is >2
     ⍵×                       ⍝ Yield the integer if so, otherwise 0
{0~⍨{                         ⍝ Remove the zeros from the resulting array

Saved 7 bytes thanks to Dennis!

Answer 9

C (gcc), 111 99 bytes

long i,x;main(a,b)char**b;{for(;++i<1L<<atol(b[1]);x>>ffsl(~x)-1&&printf("%ld,",i))x=i>>ffsl(i)-1;}

Try it online!

12 bytes shaved of thanks to @ceilingcat!

Ungolfed:

int main(int a, char **b) {
  for(long i = 0, x = 0; ++i < (1LL << atol(b[1])); ) {
    x = i >> (ffsl(i) - 1);
    if (x >> (ffsl(~x) - 1))
      printf("%ld,", i);
  }
}

The function ffsl() gives you the index of the first bit that is set in a long integer. So we loop from i = 1 to 2^number_of_bits. We set x to i shifted right until we have removed all consecutive zero bits on the least significant end. Then, we shift x right until we have removed all consecutive 1 bits at the least signficant end. If the result is still non-zero, we found a match.

Answer 10

R, 55 47 bytes

(with some help from @Alex.A)

cat(grep("10+1",R.utils::intToBin(1:2^scan())))

R doesn't have a built in function to display converted numbers in a convenient way, so I'm using R.utils::intToBin for this, while all the rest is pretty much just report the location of the matched regex expression and print to STDOUT while separated by a space.

Answer 11

CJam, 14

2qi#{2b2,#)},p

3 bytes shorter thanks to Dennis. Try it online

Answer 12

JavaScript (ES6), 69 68 67 62 bytes

a=>[...Array(1<<a).keys()].filter(i=>/01/.test(i.toString(2)))

Today I discovered a new shorter way to dynamically fill arrays without the use of fill or map. Doing x=>[...Array(x).keys()] will return an array of range 0 to x. If you want to define your own range/values, use x=>[...Array(x)].map((a,i)=>i), as it's just a few bytes longer.

Answer 13

Java, 214 165 155 154 148 141 110 bytes

This submission exploits the fact that a binary string representation of a number in Java never has a leading zero. If the string "01" appears in the binary representation of a number, that must mark the second occurrence of the number "1".

Golfed:

String f(int l){String r="";for(long i=5;i<1L<<l;++i)if(Long.toString(i,2).contains("01"))r+=i+", ";return r;}

Ungolfed:

public class NumbersWithMultipleRunsOfOnes {

  public static void main(String[] a) {
    // @formatter:off
    String[][] testData = new String[][] {
      { "3", "5" },
      { "4", "5, 9, 10, 11, 13" },
      { "5", "5, 9, 10, 11, 13, 17, 18, 19, 20, 21, 22, 23, 25, 26, 27, 29" }
    };
    // @formatter:on

    for (String[] data : testData) {
      System.out.println("Input: " + data[0]);
      System.out.println("Expected: " + data[1]);
      System.out.print("Actual:   ");
      System.out.println(new NumbersWithMultipleRunsOfOnes().f(Integer.parseInt(data[0])));
      System.out.println();
    }
  }

  // Begin golf
  String f(int l) {
    String r = "";
    for (long i = 5; i < 1L << l; ++i)
      if (Long.toString(i, 2).contains("01")) r += i + ", ";
    return r;
  }
  // End golf
}

Program output (remember, trailing delimiters are acceptable):

Input: 3
Expected: 5
Actual:   5, 

Input: 4
Expected: 5, 9, 10, 11, 13
Actual:   5, 9, 10, 11, 13, 

Input: 5
Expected: 5, 9, 10, 11, 13, 17, 18, 19, 20, 21, 22, 23, 25, 26, 27, 29
Actual:   5, 9, 10, 11, 13, 17, 18, 19, 20, 21, 22, 23, 25, 26, 27, 29, 

Answer 14

JavaScript (ES6) 76

f=n=>Array(1<<n).fill().map((_,x)=>/01/.test(x.toString(2))?x+',':'').join``

//TEST
for(i=1;i<16;i++)O.innerHTML+=i+' -> '+f(i)+'\n'

<pre id=O></pre>

Answer 15

K5, 19 bytes

This operates along similar principles as Dennis' solution, but with fewer builtins to take advantage of.

{&2<+/'~0=':'+!x#2}

First, generate a series of binary x-tuples (+!x#2), then for each tuple find every point that a digit does not match the previous if we treat the -1st element of the list as 0 for this purpose (~0=':'). Our solutions are where two is less than the sum of each run count. (&2<+/').

Showing each intermediate step is clearer:

  4#2
2 2 2 2

  !4#2
(0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1
 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1
 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1
 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1)

  +!4#2
(0 0 0 0
 0 0 0 1
 0 0 1 0
 0 0 1 1
 0 1 0 0
 0 1 0 1
 0 1 1 0
 0 1 1 1
 1 0 0 0
 1 0 0 1
 1 0 1 0
 1 0 1 1
 1 1 0 0
 1 1 0 1
 1 1 1 0
 1 1 1 1)

  ~0=':'+!4#2
(0 0 0 0
 0 0 0 1
 0 0 1 1
 0 0 1 0
 0 1 1 0
 0 1 1 1
 0 1 0 1
 0 1 0 0
 1 1 0 0
 1 1 0 1
 1 1 1 1
 1 1 1 0
 1 0 1 0
 1 0 1 1
 1 0 0 1
 1 0 0 0)

  +/'~0=':'+!4#2
0 1 2 1 2 3 2 1 2 3 4 3 2 3 2 1

  2<+/'~0=':'+!4#2
0 0 0 0 0 1 0 0 0 1 1 1 0 1 0 0

  &2<+/'~0=':'+!4#2
5 9 10 11 13

And all together:

  {&2<+/'~0=':'+!x#2}'3 4 5 
(,5
 5 9 10 11 13
 5 9 10 11 13 17 18 19 20 21 22 23 25 26 27 29)

Answer 16

Pip, 13 + 1 = 14 bytes

Takes input from the command line and uses the -s flag for spaces between output numbers.

01NTB_FI,2**a

Pretty straightforward: build range(2**a) and filter on lambda _: "01" in toBinary(_). I was pretty happy about thinking up the 01 idea independently. No quotes are needed around 01 because it scans as a numeric literal (numbers and strings are the same type in Pip).

Answer 17

Julia, 40 bytes

n->filter(i->count_ones(i$i>>1)>2,1:2^n)

This uses a somewhat different approach to the other Julia solution - rather than doing a string search for "01" in the bit string, it uses some mathematics to determine whether the number satisfies the condition.

i$i>>1 will have ones only in the places where the digit changes from zero to one, or one to zero. As such, there must be at least three ones for i to switch back and forth between zero and one enough times. count_ones finds the number of ones, and then filter removes the ones that don't have enough ones.

Answer 18

C++, 197 188 141 bytes

Note: this was written and tested using MSVC++ 2013. It appears that #includeing <iostream> includes all of the necessary C headers to make this work. It also appears that the code is no longer truly C++, but compiling using C++ allows that header trick which reduces the code size compared to including a bunch more C headers.

Using printf instead of cout also saves a couple bytes.

Golfed:

#include<iostream>
int main(int a,char**b){char c[34];for(long i=5;i<1L<<atol(b[1]);++i){_ltoa(i,c,2);if(strstr(c,"01"))printf("%ld\n",i);}}

Ungolfed:

#include <iostream>
int main(int a, char **b) {
  char c[34];
  for (long i = 5; i < 1L << atol(b[1]); ++i) {
    _ltoa(i, c, 2);
    if (strstr(c, "01"))
      printf("%ld\n", i);
  }
}

Answer 19

Perl 5, 55 53 49 47 41 bytes

sprintf("%b",$_)=~/01/&&say for 0..2**<>

54 52 48 46 40 bytes, plus one for the -E flag instead of -e.

---

Thanks to xnor for the hint about using /01/ instead of /10+1/, which saved two bytes.

Thanks to Dennis for the advice to use <> instead of $ARGV[0], which saved six bytes.

Answer 20

C, 84 81 bytes

long i,j,k;main(){for(scanf("%ld",&j);++i<1L<<j;k&k+1&&printf("%ld ",i))k=i|i-1;}

This is based on the comments I made on another C answer to this question about the possibility of using simple bitwise operators. It works by switching all trailing 0 bits to 1 in the statement i|(i-1). Then it switches all trailing 1 bits to 0 using k&(k+1). This will result in a zero if there is only one run of ones and non-zero otherwise. I do make the assumption that long is 64-bit but could correct this at the expense of three bytes by using int64_t instead.

Ungolfed

long i,j,k;
main()
{
    for(scanf("%ld",&j);++i<1L<<j;)
    {
        k=i|(i-1);
        if((k&(k+1)) == 0)
            printf("%d ",i);
    }
}

Answer 21

Pyth, 20 17 16 bytes

f:.BT"10+1"ZU^2Q

Live demo.

Answer 22

Python 2.7, 89 bytes

print[i for i in range(1,2**input())if[n[:1]for n in bin(i)[2:].split("0")].count("1")-1]

I think this could be golfed a bit.

Answer 23

TI-BASIC, 34 32 30 bytes

For a TI-83+/84+ series calculator.

For(X,5,e^(Ans
If log(sum(2=int(4fPart(X/2^randIntNoRep(1,Ans
Disp X
End

For a number to contain two runs of 1s, it must contain two 10s when we tack a trailing zero onto the binary representation.

Rather than generate the binary representation and check for a 10, we test pairs of bits mathematically by using remainder by 4 (int(4fPart(), which will give 2 where there is a 10. Because we don't care about order, randIntNoRep( is the shortest way to generate the list of exponents.

We use log( to check for the number of runs:

• If there are at least 2 runs, then the log( is positive, and the number is displayed.

• If there is one run, then the log( is 0, and the number is not displayed.

• If there are no runs (which first happens at X=2^Ans), then log( throws an ERR:DOMAIN, stopping the output at exactly the right point.

We use e^(Ans as the ending argument of the For( loop—it's always greater than 2^Ans, but e^( is a single token, so it's one byte shorter.

Input/output for N=4:

4:prgmRUNSONES
               5
               9
              10
              11
              13

Then the calculator throws an error; the error screen looks like this:

ERR:DOMAIN
1:Quit
2:Goto

When 1 is pressed, the home screen is displayed again:

4:prgmRUNSONES
               5
               9
              10
              11
              13
           Error

TI calculators store all numbers in a BCD float with 14 digits of precision, not an int or binary float. Therefore, divisions by powers of two greater than 2^14 may not be exact. While I have verified that the trickiest numbers, 3*2^30-1 and 2^32-1, are handled correctly, I have not ruled out the possibility of rounding errors. I would however be surprised if there were errors for any input.

Answer 24

• this doesnt beat flawr's answer but i couldnt resist the charm of the question

matlab(90)(70)

j=input('');for l=2:j-1,a=1;for k=l:-1:2,a=a+2^k;a:a+2^(k-1)-2,end,end

execution

4

ans =

5

ans =

9    10    11

ans =

13

principle

• The series of numbers are a result of consequent strip of 1's, which does mean f(n,l)=2^l+2^(l+1)+....2^n

Any number taken from the interval ]f(n,l),f(n,l)+2^(l-1)[ where l>1 verifies this condition, so the outcome is a result of the negation of this series in terms of n.

x=1

x=x+1=01,

x=x+2^0=11,

x=x+1=001,

x=x+2^1=011,

x=x+2^0=111,

x=x+1=0001,

x=x+2^2=0011,

x=x+2^1=0111,

x=x+2^0=0111,

x=x+1=1111 ...

x+1, x=x+2^n, x=x+2^(n-1)...x=x+2^0

My program prints the range between each two lines (if exists)

---

Edit: unfortunately that doesnt make it golfed more but i wanted to add another approach of proceding this idea

after a period of struggle i succeded to find a mathematical representation for this series which is:

2^l(0+1+2^1+...2^k) with l+k < n

=2^l(2^k-1)

score=90

@(n)setdiff(0:2^n-1,arrayfun(@(x)2^mod(x,(n+1)-fix(x/(n+1)))*(2^fix(x/(n+1))-1),0:(n+1)^2))

Answer 25

C, 103 102 bytes

long i,x;main(int a,char**b){for(;++i<1L<<atoi(b[1]);)for(x=i;x>4&&(x%4!=1||!printf("%ld,",i));x/=2);}

Expanding (actually contracting) on G.Sliepen entry, taking advantage of xnor remark on the 01 pattern in the binary representation, but using only standard functions and some bit twiddling.

Ungolfed version:

long i, x;
main(int a, char**b) {
    for (; ++i < 1L << atoi(b[1]);) {
        for (x = i; x > 4 && (x % 4 != 1 || !printf("%ld,", i)); x /= 2)
            ;
    }
}

The inner loop scans i for the binary pattern 01 by iteratively shifting x to the right as long as it has 3 bits left. printf returns the number of characters printed, hence never 0, so the inner loop test fails after the printf, avoiding the need for a break statement.

C++, 129 128 bytes

Adapting the same idea, the C++ variant is here:

#include<iostream>
long i,x;int main(int a,char**b){for(;++i<1L<<atoi(b[1]);)for(x=i;x>4&&(x%4!=1||!(std::cout<<i<<','));x/=2);}

Technically, I should make i a long long to ensure 64 bit operation and compute upto 2^32 for an extra 5 bytes, but modern platforms have 64 bit ints.

Answer 26

JavaScript ES6, 60 bytes

Code

n=>[...Array(1<<n).keys()].filter(g=x=>x>4?x%4==1|g(x>>1):0)

Try it online!

Explanation

[...Array(1<<n).keys()]                                          Create an array of numbers from 0 to 2^n-1
                       .filter(                                  Find the numbers which meet criteria
                               g=x=>x>4?                  :0     If x is less than or equal to four return zero (false/invalid)
                                        x>4?x%4==1|              If the binary pattern ends with 01 return one (true/valid)
                                                   g(x>>1)       Otherwise bitshift right by one and try again

Answer 27

PowerShell, 80 bytes

while([Math]::Pow(2,$args[0])-gt$n++){$n|?{[Convert]::ToString($n,2)-match"01"}}

Answer 28

Python, 44 Bytes

Okay, this could probably be shorter but it's my first codegolf:

x=1
while True:
    if '01' in bin(x):
        print(x)
    x+=1

Think this answers the question, please don't down vote if it doesn't, just post whats wrong with it below.

Answer 29

another different matlab answer of good score.

matlab 60(57)

@(n)find(mod(log2(bitcmp(1:2^n,fix(log2(1:2^n)+1))+1),1))

execution

>> @(n)find(mod(log2(bitcmp(1:2^n,fix(log2(1:2^n)+1))+1),1))

ans =

@(n)find(mod(log2(bitcmp(1:2^n,fix(log2(1:2^n)+1))+1),1))

>> ans(5)

ans =

 5     9    10    11    13    17    18    19    20    21    22    23    25    26    27    29

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• The idea is picking up numbers x where the binary representation of -(x)+1 doesnt contain only one occurence of 1

example:

0000111 is rejected because ~x=1111,~x+1=00001 contains one digit=1

0100111 is accepted because ~x=1011,~x+1=0111 contains many 1's

Answer 30

k4, 32 28 bytes

{&1<+/'1='-':'0b\:'o:!*/x#2}

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explanation

                      */x#2   /x^2
                   o:!        /assign o to [0,x^2 -1), e.g. for x=2, o<-0 1 2 3
              0b\:'           /get binary value of each array elem
          -':'                /get delta of adjacent digits in each binary num
    +/'1='                    /sum each array to find how many deltas=1
 &1<                          /return elems where there were greater than 1 deltas=1